The question is incomplete, here is the complete question:
A sample of gas weighs 10.1 g and occupies a volume of 5.65 L at 60 °C and 580 torr. Identify the gas sample.
(a)
(molar mass = 119.4 g/mol)
(b)
(molar mass = 44.02 g/mol)
(c)
(molar mass = 64.07 g/mol)
(d)
(molar mass = 17.03 g/mol)
(e)
(molar mass = 70.90 g/mol)
<u>Answer:</u> The gas sample is
having molar mass of 64 g/mol
<u>Explanation:</u>
To calculate the molar mass of the gas, we use the equation given by ideal gas equation:
PV = nRT
Or,

where,
P = Pressure of the gas = 580 torr
V = Volume of the gas = 5.65 L
w = Weight of the gas = 10.1 g
M = Molar mass of gas = ?
R = Gas constant = 
T = Temperature of the gas = ![60^oC=[60+273]K=333K](https://tex.z-dn.net/?f=60%5EoC%3D%5B60%2B273%5DK%3D333K)
Putting values in above equation, we get:

Hence, the gas sample is
having molar mass of 64 g/mol